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\markboth{\hfil Dirichlet problem for quasi-linear elliptic equations
\hfil EJDE--2002/82}
{EJDE--2002/82\hfil Azeddine Baalal \& Nedra BelHaj Rhouma \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 82, pp. 1--18. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
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Dirichlet problem for quasi-linear elliptic equations
%
\thanks{ {\em Mathematics Subject Classifications:} 31C15, 35B65, 35J60.
\hfil\break\indent
{\em Key words:} Supersolution, Dirichlet problem, obstacle problem,
nonlinear potential theory.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted April 9, 2002. Published October 2, 2002. \hfil\break\indent
Supported by Grant DGRST-E02/C15 from Tunisian Ministry of
Higher Education.} }
\date{}
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\author{Azeddine Baalal \& Nedra BelHaj Rhouma}
\maketitle
\begin{abstract}
We study the Dirichlet Problem associated to the quasilinear
elliptic problem
\begin{equation*}
-\sum_{i=1}^{n}\frac{\partial }{\partial x_i}\mathcal{A}_i(x,u(x),
\nabla u(x))+\mathcal{B}(x,u(x),\nabla u(x))=0.
\end{equation*}
Then we define a potential theory related to this problem and we
show that the sheaf of continuous solutions satisfies the Bauer
axiomatic theory.
\end{abstract}
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}{Remark}[section]
\section{Introduction}
The objective of this paper is to study the weak solutions of
the following quasi-linear elliptic equation in $\mathbb{R}^{d}$,
($d\geq 2$):
\begin{equation}
-\sum_{i=1}^{n}\frac{\partial }{\partial x_i}\mathcal{A}_i(x,u(x),
\nabla u(x))+\mathcal{B}(x,u(x),\nabla u(x))=0\quad \label{eq1}
\end{equation}
where $\mathcal{A}_i:\mathbb{R}^{d}\times \mathbb{R}\times \mathbb{R}
^{d}\to \mathbb{R}$ and $\mathcal{B}:\mathbb{R}^{d}\times \mathbb{R}\times
\mathbb{R}^{d}\to \mathbb{R}$ are given Carath\'{e}odory functions
satisfying the conditions introduced in section 2.
An example of equation (\ref{eq1}) is the perturbed $p$-Laplace
equation
\begin{equation}
-\mathop{\rm div}(|\nabla u| ^{p-2}\nabla u)+\mathcal{B}(.,u,\nabla
u)=0, \quad 1
0$,
if $\xi \neq \xi '$.
\item[(P3)] $ \mathcal{A}(x,\zeta,\xi )\xi \geq \alpha | \xi|^p-d_0(x)
| \zeta | ^p-e(x)$
\item[(P4)] $ | \mathcal{B}(x,\zeta,\xi )| \leq k(x)+b(x)|
\zeta | ^{\alpha }+c| \xi | ^{r}$,
$0v\}$,
$\Omega _{2}=\{ x\in \Omega :u\leq v\}$ and put
$I=\int_{\Omega }\mathcal{A}(.,u\vee v,\nabla (u\vee v))\nabla \varphi
=I_{1}+I_{2}$ where
$$
I_{1}=\int_{\Omega _{1}}\mathcal{A}(.,u,\nabla u)\nabla
\varphi \mbox{ \ and \ }I_{2}=\int_{\Omega
_{2}}\mathcal{A}(.,v,\nabla v)\nabla \varphi .
$$
Let $\rho _{n}:\mathbb{R}\to \mathbb{R}$ be such that
$\rho _{n}\in \mathcal{C}^{1}(\mathbb{R})$,
$$
\rho _{n}(t)=\begin{cases}
1 & \mbox{if } t \geq 1/n \\
0 & \mbox{if } t \leq 0 \end{cases}
$$
and $\rho _{n}'>0$ on $] 0,1/n[$. For each $x\in \Omega$ define
$q_{n}(x)=\rho _{n}((u-v)(x))$. We see that
$q_{n}\in W_{{\rm loc}}^{1,p}(\Omega )$, $q_{n}\to 1_{\Omega _{1}}$ and
$\| q_{n}\|_{\infty }\leq 1$. It follows by Lebesgue's Theorem of dominated
convergence that $I_{1}={\lim_{{n\to \infty }} }\int_{\Omega _{1}}q_{n}
\mathcal{A}(.,u,\nabla u).\nabla \varphi $ and
$I_{2}={\lim_{n\to \infty }}
\int_{\Omega_{2}}(1-q_{n})\mathcal{A}(.,v,\nabla v).\nabla \varphi $. Hence
\begin{eqnarray*}
\int_{\Omega }q_{n}\mathcal{A}(.,u,\nabla u).\nabla \varphi
&=&\int_{\Omega }\mathcal{A}(.,u,\nabla u)\nabla
.(q_{n}\varphi )-\int_{\Omega }\mathcal{A}(.,u,\nabla
u)\varphi .\nabla (q_{n}) \\ \ &\leq &-\int_{\Omega
}\mathcal{B}(.,u,\nabla u)(q_{n}\varphi )-\int_{\Omega
_{n}}\mathcal{A}(.,u,\nabla u)\varphi .\nabla (q_{n}),
\end{eqnarray*}
where $\Omega _{n}=\{ x\in \Omega :v0 $ and every supersolution (resp. subsolution) $u$ of
(\ref{eq1}), the function
$u+k$ (resp. $u-k$) is also a supersolution (resp. subsolution) of
(\ref{eq1})
\begin{remark} \label{rm1.1}
1) Suppose that for each $u\in W_{{\rm loc}}^{1,p}(\Omega )$ and each
$k>0$,
\begin{equation}
\int (\mathcal{A}(.,u+k,\nabla u)-\mathcal{A}(.,u,\nabla u)).\nabla \varphi
+\int (\mathcal{B}(.,u+k,\nabla u)-\mathcal{B}(.,u,\nabla u))\varphi \geq 0
\label{SH}
\end{equation}
for every nonnegative function $\varphi \in W_{0}^{1,p}(\Omega )$.
Then $\mathcal{L}$ satisfies the property ($\pm $).
\noindent 2) Note that if $\mathcal{L}(u)=-\sum_{j}\frac{\partial}{\partial
x_{j}}(\sum_ia_{ij}\frac{\partial u}{\partial x_i}
+d_{j}u)+(\sum_ib_i\frac{\partial u}{\partial x_i}+cu)$ is a
linear elliptic operator of second order satisfying the conditions of
\cite{Her}, then (\ref{SH}) is equivalent to
$(-\sum_{j}(d_{j})+c)\geq 0$ in the distributional sense.
\noindent 3) Suppose that $\mathcal{A}(x,\zeta ,\xi )=\mathcal{A}(x,\xi )$
and for a.e. $x\in \Omega $ and $\xi \in \mathbb{R}^{d}$ the map:
$\zeta \to \mathcal{B}(x,\zeta ,\xi )$ is increasing. Then the property
($\pm $) holds.
\end{remark}
\section{Comparison principle} \label{cp}
In this section, we will give some conditions needed for the comparison
principle. This principle makes it possible to solve the Dirichlet problem
and to develop a potential theory in our case.
We say that the \emph{comparison principle} holds for $\mathcal{L}$, if for
every supersolution $u$ and every subsolution $v$ of (\ref{eq1}) on $\Omega $%
, such that
$$ \limsup_{x\to y} v(x)\leq \liminf_{x\to y} u(x)
$$
for all $y\in \partial \Omega $ and both sides of the inequality are not
simultaneously $+\infty $ or $-\infty $, we have $v\leq u$ \ a.e. in
$\Omega $.
\begin{theorem} \label{thm1}
Suppose that the operator $\mathcal{L}$ satisfies
either one of the property \emph{(}$\pm $\emph{)} and the following
strict monotony condition (see \cite{Ne}):
$$
(\mathcal{A}(x,\zeta ,\xi )-\mathcal{A}(x,\zeta ',\xi
')).(\xi -\xi ')+(\mathcal{B}(x,\zeta ,\xi
)-\mathcal{B}(x,\zeta ',\xi '))(\zeta -\zeta
')>0 $$ for $(\zeta ,\xi )\neq (\zeta ',\xi
')$. Let $u$ be a supersolution and $v$ be a subsolution
of \emph{(\ref{eq1})}, on $\Omega $, such that $$
{\limsup_{x\to y} }v(x)\leq { \liminf_{x\to y}
}u(x)
$$
for all $y\in \partial \Omega $ and both sides of the inequality are not
simultaneously $+\infty $ or $-\infty $, then $v\leq u$ \ a.e. in $%
\Omega .$
\end{theorem}
\paragraph{Proof.}
Let $\varepsilon >0$ and $K$ be a compact subset of $\Omega $ such that $%
v-u\leq \varepsilon $ on $\Omega \backslash K$, then the function $%
\varphi =(v-u-\varepsilon )^{+}$ $\in W_{0}^{1,p}(\Omega )$. Testing by $%
\varphi $, we obtain that
\begin{eqnarray*}
0 & \leq & \int_{ v>u+\varepsilon}(\mathcal{A}
(.,u+\varepsilon ,\nabla u)-\mathcal{A}(.,v,\nabla
v))\nabla (v-u-\varepsilon ) \\
&& + \int_{v>u+\varepsilon} (\mathcal{B} (.,u+\varepsilon ,\nabla
u)-\mathcal{B}(.,v,\nabla v))(v-u-\varepsilon ) \leq 0\,.
\end{eqnarray*}
Hence $\nabla (v-u-\varepsilon )^{+}=0$ and $(v-u-\varepsilon )^{+}$ $=0$
a.e. in $\Omega $. It follows that $v\leq u+\varepsilon $ a.e. in
$\Omega $ and therefore $v\leq u$ a.e. in $\Omega $
\hfill$\square$
\begin{corollary} \label{cor1}
we suppose that $\mathcal{A}(x,\zeta ,\xi )=\mathcal{A}(x,\xi )$
and $\mathcal{B}(x,\zeta ,\xi )=\mathcal{B}(\zeta )$ such that the map $%
\zeta \to \mathcal{B}(x,\zeta )$ is increasing \ for a.e. $x$ in $%
\Omega $. Then, the comparison principle holds.
\end{corollary}
\begin{theorem}
Suppose that
\begin{description}
\item[i)] $[ \mathcal{A}(x,\zeta ,\xi )-\mathcal{A}(x,\zeta
',\xi ')].(\xi -\xi ')\geq \gamma | \xi -\xi '| ^p$ for all
$\zeta ,\zeta '$ in $\mathbb{R}$, for all $\xi ,\xi '\in \mathbb{R}^{d}$,
a.e. $x$ in $ \Omega $ and for some $\gamma>0$ .
\item[ii)] For a.e. $x\in \Omega $ and for all $\xi \in \mathbb{R}^{d}$, the
map $\zeta \to \mathcal{B}(x,\zeta ,\xi )$ is increasing,
\item[iii)] $| (\mathcal{B}(x,\zeta ,\xi )-\mathcal{B}(x,\zeta ,\xi
')| \leq b(x,\zeta )| \xi -\xi '| ^{p-1}$ for a.e. $x\in \Omega $,
for all
$\zeta \in \mathbb{R}$ and for all $\xi ,\xi '\in \mathbb{R}^{d}$.
Where $\sup_{| \zeta | \leq M}b(.,\zeta )\in
L_{{\rm loc}}^{s}(\Omega )$, $s>d$, for all $M>0$.
\end{description}
Then the comparison principle holds.
\end{theorem}
\paragraph{Proof.} The main idea in this proof comes from
Professor J. Maly'.
Let $\rho >0$, $M=\sup (v-u)$ and put $w=v-u-\rho $. Take $w^{+}$
as test function . Then, we get $$ \int_{\Omega }\left[
\mathcal{A}(.,u,\nabla u)-\mathcal{A}(.,v,\nabla v)\right] .\nabla
(w^{+})+\int_{\Omega }\left[ \mathcal{B}(.,u,\nabla
u)-\mathcal{B}(.,v,\nabla v)\right] (w^{+})\geq 0
$$
and by consequence
\begin{eqnarray*}
\gamma \int_{\Omega }| \nabla w^{+}| ^p &\leq
&\int_{\Omega }b(x,v)| \nabla w^{+}| ^{p-1}w^{+}
\\
&\leq &C\Big[ \int_{\Omega }| \nabla w^{+}| ^p%
\Big] ^{\frac{p-1}{p}}\Big[ \int_{\Omega }( w^{+})
^{p^{\ast }}\Big] ^{\frac{1}{p^{\ast }}}| A_{\rho }| ^{\frac{s-d}{sd}} \\
&\leq &C\| \nabla w^{+}\| _{p}^p\, | A_{\rho }| ^{\frac{s-d}{sd}}.
\end{eqnarray*}
where $A_{\rho }=\{ \rho 0$. Thus, $v\leq u$ on $\Omega $ \hfill$\square$
\section{Dirichlet Problem}
\subsection*{Existence of solutions for $0\leq \alpha \leq p-1$ and $0\leq r\leq p-1$}
\paragraph{Definition} %def3.1
Let $g\in W^{1,p}(\Omega )$. We say that $u$ is a solution of problem
$(P)$ if
\begin{gather*}
u-g\in W_{0}^{1,p}(\Omega ), \\
\int_{\Omega }\mathcal{A}(.,u,\nabla u).\nabla \varphi
+\int_{\Omega }\mathcal{B}(.,u,\nabla u)\varphi =0\quad \forall
\varphi \in W_{0}^{1,p}(\Omega ).
\end{gather*}
\begin{remark} %\label{rem1}
Put $v=u-g$, then $u$ is a solution of the above problem $(P)$ if
and only if $v$ is a solution of
\begin{equation}
\begin{gathered}
u\in W_{0}^{1,p}(\Omega ) \\
\int_{\Omega}{\mathcal{A}_{g}}(.,u,\nabla u)\nabla \varphi
+\int_{\Omega }{\mathcal{B}_{g}}(.,u,\nabla u)\varphi
=0,\quad \forall \varphi \in W_{0}^{1,p}(\Omega ),
\end{gathered} \label{P'}
\end{equation}
where ${\mathcal{A}_{g}}(.,u,\nabla u)=\mathcal{A}(.,u+g,\nabla (u+g))$ and
${\mathcal{B}_{g}}(.,u,\nabla u)=\mathcal{B}(.,u+g,\nabla (u+g))$.
\end{remark}
Let $T: W_{0}^{1,p}(\Omega ) \to W_{0}^{-1,p'}(\Omega )$
be the operator defined by
$$ \langle T(u),v\rangle =\int {\mathcal{A}_{g}}(.,u,\nabla u)\nabla v
+\int {\mathcal{B}_{g}}(.,u,\nabla u)v
\quad \forall v\in W_{0}^{1,p}(\Omega). $$ Next we will establish
the existence of solution of \eqref{P'} when $0\leq \alpha \leq
p-1$ and $0\leq r\leq p-1$. Let $C=C(d,p)$ be a constant such that
$\| u\|_{p*}\leq C \| u\|_{p}$ for every $u\in
W_{0}^{1,p}(\Omega)$. Then, we get the following result.
\begin{proposition} \label{prop0}
Suppose that $0\leq \alpha \leq p-1$ and $0\leq r\leq
p-1$. If $\Omega $ is small (i.e $\alpha >C(\| d_{0}\|
_{n/p}+\| b\| _{n/p})$), then the operator $T$ is coercive.
\end{proposition}
\paragraph{Proof.}
We have
\begin{eqnarray*}
\langle T(u),u\rangle & = &
\int \mathcal{A}(u+g,\nabla (u+g))\nabla u+\int \mathcal{B}(u+g,\nabla (u+g))u \\
& \geq & \big(\alpha -C\| d_{0}\| _{d/p}
-C\| b\| _{d/p}\big)\| \nabla u\| _{p}^p-H_{1}
(\|u\| ,\| \nabla u\| ,\| g\| ,\|
\nabla g\| )
\end{eqnarray*}
where $C=C(d,p)$ and the growth of $H_{1}$ in $\| u\| $ and $\| \nabla u\| $
is less then $p-1$. So, let $\Omega $ be small enough such that
$\alpha >C(\| d_{0}\|_{n/p}+\|b\| _{n/p})$.
Hence, $\frac{\langle T(u),u\rangle }{\| \nabla u\| _{p}}\to +\infty $ as
$\| \nabla u\| _{p}\to +\infty $ and therefore the operator $T$ is
coercive. \hfill$\square$
\begin{proposition} \label{lem1}
Suppose that $0\leq \alpha \leq p-1$ and $0\leq r\leq
p-1$. Then, the operator $T$ is pseudomonotone and satisfies the
well known property ($S_{+}$):\\
If $u_{n}\rightharpoonup u$ and
$\limsup_{n\to \infty }\langle T(u_{n})-T(u),u_{n}-u\rangle \leq 0 $,
then $ u_{n}\to u$.
\end{proposition}
The proof of this proposition is found in \cite{Ma}.
\begin{theorem}\label{thmoo}
Suppose that $T$ satisfies the coercive condition on $\Omega $.
Then \eqref{P'} has at least one weak solution in
$W_{0}^{1,p}(\Omega )$.
\end{theorem}
\paragraph{Proof.}
The operator $T$ is pseudomonotone, bounded continuous and coercive. Hence,
by \cite{Ne} $T$ is surjective. \hfill$\square$
\subsection*{Existence of solutions for $\alpha\geq 0$ and $p-1p^*{}'$
such that for all
$\xi \in \mathbb{R}^{d}$ and all $\zeta$ with $ \varphi
(x)\leq \zeta \leq \psi (x)$, $| \mathcal{B} (x,\zeta ,\xi
)| \leq k(x)+c| \xi | ^{r}$ a.e.$x\in \Omega$.
Then, \eqref{sD} has at least one solution $u\in
W_{0}^{1,p}(\Omega )$ such that $\varphi \leq u\leq \psi $.
\end{theorem}
\begin{proposition}
Suppose that \eqref{sD} admits a pair of bounded lower
subsolution $u$ and upper supersolution $v$ such that $u\leq v$,
then there exists a solution $w$ of \eqref {sD} such that
$u\leq w\leq v$.
\end{proposition}
\paragraph{Proof.}
Let $M$ be a positive real such that
$\| u\| _{\infty},\| v\| _{\infty },\| g\| _{\infty }\leq M$. Then, for each
$\zeta $ such that $u(x)-g(x)\leq \zeta \leq v(x)-g(x)$, we have
$| \mathcal{B}(x,\zeta ,\xi )| \leq k(x)+b(x)M^{\alpha}+2^{r}c| \nabla g| ^{r}
+c| \xi | ^{r}$ for a.e. $x\in \Omega $. In addition, $u$
(resp. $v$) is a lower subsolution (resp. upper supersolution) of
(\ref{sD}). Hence by the last Theorem, there exists a solution $w$ of
(\ref{sD}) such that $u\leq w\leq v$. \hfill$\square$
\begin{corollary}
Suppose that all positive constants are supersolutions and all negative
constants are subsolutions. Then for each $g\in W^{1,p}(\overline{\Omega }%
)\cap L^{\infty }(\Omega )$, there exists a bounded solution $w$
of (\ref{sD}) such that $\| w\| _{\infty }\leq \|
g\| _{\infty }$.
\end{corollary}
\paragraph{Proof.} We see that $v=\| g\| _{\infty }$ is an upper
supersolution and $u=-\| g\| _{\infty }$ is a lower
subsolution. Hence by the Proposition given above, we get a solution
$u\leq w\leq v$ \hfill$\square$
\subsection{Dirichlet Problem}\label{sDP}
In this section, we assume that $\mathcal{A}(.,0,0)=0$ and
$\mathcal{B}(.,0,0)=0$ a.e. in $\Omega $, that the property ($\pm $)
is satisfied, and that the comparison principle holds.
Suppose that the open set $\Omega $ is regular ($p-$regular)
\cite{Ma,Hei}. Then it is known that if $u$ is a solution of
(\ref{eq1}) in $\Omega $ satisfying $u-f\in W_{0}^{1,p}(\Omega )$
with $f\in W^{1,p}(\Omega )\cap C(\overline{\Omega })$, then $$
\lim_{x\to z} u(x)=f(z) \quad \forall z\in \partial \Omega . $$
\paragraph{Definition} %def3.1
Let $f$ be a continuous function on $\partial \Omega $. We say
that $u$ $\in C(\overline{\Omega })\cap $ $W_{{\rm loc}}^{1,p}(\Omega )$
solves the Dirichlet problem with boundary value $f$ if $u$ is a
solution of \eqref{eq1} such that $\lim_{x\to z}u(x)=f(z)$,
for all $z\in \partial \Omega $.
\begin{theorem} \label{thDP}
For each $f\in C(\partial \Omega )$, there exists $u$ in
$C(\overline{\Omega })\cap $ $W_{{\rm loc}}^{1,p}(\Omega )$
solving the Dirichlet problem with boundary value $f$.
\end{theorem}
\paragraph{Proof}
By the Tieze's extension Theorem, we can assume that
$f\in C_{c}^{\infty }(\mathbb{R}^{d})$. Let $(f_{n})_{n}$ be a sequence
of mollifiers of $f$ such that $\| f_{n}-f\| \leq 1/2^n$ on
$\overline{\Omega}$ .
let $u_{n}$ denote the continuous solution of
\begin{equation}
\begin{gathered}
u_{n}-f_{n}\in W_{0}^{1,p}(\Omega ), \\
\int_{\Omega}\mathcal{A}(.,u_{n},\nabla u_{n})\nabla \varphi
+\int_{\Omega }\mathcal{B}(.,u_{n},\nabla u_{n})\varphi =0,
\quad \forall \varphi \in W_{0}^{1,p}(\Omega ).
\end{gathered} \label{En}
\end{equation}
So, by the comparison principle, $|u_{n}-u_{m}| \leq \frac{1}{2^{n}}+\frac{1}{2^{m}}$. Hence,
the sequence $(u_{n})_{n}$ converges uniformly on
$\overline{\Omega }$ to a continuous function $u$. Let $M$ be a
positive real such that for all $n$: $| f_{n}|+| f| \leq M$ and
$| u_{n}| +| u|\leq M$ on $\Omega $.
Let $G\subset \overline{G}\subset \Omega $ , take $\varphi $ as a
test function in (\ref{En}) such that $\varphi =\eta
^pu_{n},\eta \in C_{c}^{\infty }(\Omega ),0\leq \eta \leq 1$ and
$\eta =1$ on $G$. Then
\begin{multline*}
\int_{\Omega}\mathcal{A}(.,u_{n},\nabla u_{n})\eta ^p\nabla(u_{n})\\
=-p\int_{\Omega }\mathcal{A}(.,u_{n},\nabla
u_{n})u_{n}\eta ^{p-1}\nabla (\eta )-\int_{\Omega
}\mathcal{B}(.,u_{n},\nabla u_{n})u_{n}\eta ^p
\end{multline*}
Using the assumptions on $\mathcal{A}$ and $\mathcal{B}$, we get
\begin{eqnarray*}
\lefteqn{\alpha \int_{\Omega }\eta ^p| \nabla (u_{n})|^p }\\
&\leq &pM\int_{\Omega }k_{0}| \nabla \eta |
+pM^p\int_{\Omega }b_{0}| \nabla \eta | +
pM\int_{\Omega }a| \nabla u_{n}| ^{p-1}\eta
^{p-1}| \nabla \eta | \\ &&\ +cM\int_{\Omega
}| \nabla u_{n}| ^{r}\eta ^p+\int_{\Omega
}(M^pd_{0}+Mk+M^{\alpha +1}b+e) \\
&\leq & a(p-1)^{-1}M\varepsilon ^{\frac{p}{p-1}}(\int_{\Omega
}| \nabla u_{n}| ^p\eta ^p)+crp^{-1}M\varepsilon ^{\frac{p}{r}}
(\int_{\Omega }| \nabla u_{n}| ^p\eta ^p)\\
&&+C(M,\Omega ,\eta ,\nabla \eta ).
\end{eqnarray*}
Thus, for $\varepsilon $ small enough, we obtain
$$
\int_{G}| \nabla (u_{n})| ^p\leq C(M,\Omega
,\eta ,\nabla \eta ,\varepsilon ).
$$
So $(\nabla u_{n})_{n}$ is bounded in $L^p(G)$ and therefore
$(\nabla u_{n})_{n}$ converges weakly to $\nabla u$ in
$(L^p(G))^{d}$.
Fix $D$ an open subset of $G$ and let $\eta \in C_{0}^{\infty
}(G)$ such that $0\leq \eta \leq 1$ and $\eta =1$ on $D$. Take
$\psi =\eta (u_{n}-u)$ as test function, then
\begin{eqnarray*}
\lefteqn{-\int_{\Omega }\eta \mathcal{A}(.,u_{n},\nabla u_{n}).\nabla
(u_{n}-u)}\\
&=&\int_{\Omega }(u_{n}-u)\mathcal{A}(.,u_{n},\nabla
u_{n}).\nabla \eta +
\int_{\Omega }\mathcal{B}(.,u_{n},\nabla u_{n})(u_{n}-u)\eta
\end{eqnarray*}
Since $\mathcal{A}(.,u_{n},\nabla u_{n})$ is bounded in
$L^{p'}(G)$ and $\mathcal{B}(.,u_{n},\nabla u_{n})$ is
bounded in $L^{q}(G)$,
\begin{gather*}
{\lim_{n\to \infty } }\int_{G}\mathcal{A}
(.,u_{n},\nabla u_{n})(u_{n}-u)\nabla \eta =0,\\
\lim_{n\to \infty } \int_{G}\mathcal{B}
(.,u_{n},\nabla u_{n})(u_{n}-u) \eta =0.
\end{gather*}
Consequently,
$\lim_{n\to \infty } \int_{G}\mathcal{A}(.,u_{n},\nabla u_{n})\eta\nabla
(u_{n}-u)=0$ and
$$
\lim_{n\to \infty } \int_{G} (\mathcal{A}(.,u_{n},\nabla u_{n})
-\mathcal{A}(.,u_{n},\nabla u))\nabla (u_{n}-u)=0.
$$
To complete the proof, we need to prove that $(\nabla u_{n})_{n}$
converges to $\nabla u$ a.e. in $\Omega $. That is the aim of the
following lemma.
\begin{lemma} \label{lweak}
Let $G\subset \Omega $ and suppose that the sequence
$(\nabla u_{n})_{n}$ is bounded in $L^p(G)$ and
$$
{\lim_{n\to \infty } }\int_{G}\left[ \mathcal{A}
(.,u_{n},\nabla u_{n})-\mathcal{A}(.,u,\nabla u)\right] .\nabla
(u_{n}-u)=0.
$$
Then $ \mathcal{A}(.,u_{n},\nabla u_{n})\to
\mathcal{A}(.,u,\nabla u)$ weakly in $L^{p'}(G)$.
\end{lemma}
\paragraph{Proof.}
Put $v_{n}=\left[ \mathcal{A}(.,u_{n},\nabla u_{n})-\mathcal{A}%
(.,u_{n},\nabla u)\right] .\nabla (u_{n}-u)$. Since
\begin{eqnarray*}
\int_{G}v_{n} &=&\int_{G}\left[ \mathcal{A}(.,u_{n},\nabla
u_{n})-\mathcal{A}(.,u,\nabla u)\right] .\nabla (u_{n}-u) \\
&&-\int_{G}\left[ \mathcal{A}(.,u_{n},\nabla u)-\mathcal{A}%
(.,u,\nabla u)\right] .\nabla (u_{n}-u),
\end{eqnarray*}
for a subsequence we get
$$ {\lim_{n\to \infty }
}\left[ \mathcal{A}(.,u_{n},\nabla
u_{n})-\mathcal{A}(.,u_{n},\nabla u)\right] .\nabla (u_{n}-u)=0 $$
a.e. $x\in G\setminus N$ with $| N| =0$. Let $x\in
G\setminus N$. By the assumptions on $\mathcal{A}$ we have
$$
v_n(x)\geq \alpha | \nabla u_n(x)| ^p-F(|
\nabla u_n(x)| ^{p-1},| \nabla u(x)| ^{p-1}).
$$
Consequently, $(\nabla u_{n}(x))_{n}$ is bounded and converges to
some $\xi \in\mathbb{R}^{d}$. It follows that
$[\mathcal{A}(.,u,\xi )-\mathcal{A}(.,u,\nabla u)].(\xi -\nabla u)=0$
and hence $\xi =\nabla u$. Finally we conclude that
$\mathcal{A}(.,u_{n},\nabla u_{n})\to \mathcal{A}(.,u,\nabla u)$ a.e. in
$G$ and $\mathcal{A}(.,u_{n},\nabla u_{n})$ converge weakly to
$\mathcal{A}(.,u,\nabla u)$ in $L^{p'}(G)$. \hfill$\square$
Now we go back to the proof of Theorem \ref{thDP}.
Using Lemma \ref{lweak}, we conclude that
$\nabla u_{n}\to \nabla u$ a.e. in $\Omega $ and
$\mathcal{A}(.,u_{n},\nabla u_{n})\rightharpoonup \mathcal{A}(.,u,\nabla u)$
in $L^{p'}(D)$. Hence,
$$
\int_{D}\mathcal{A}(.,u,\nabla u)\nabla \varphi +\int_{D}%
\mathcal{B}(.,u,\nabla u)\varphi =0
\quad \forall \varphi \in C_{0}^{\infty }(\Omega ).
$$
Moreover, using the fact that
$$
-\frac{1}{2^{n}}-\frac{1}{2^{m}}\leq u_{m}-u_{n}\leq \frac{1}{2^{n}%
}+\frac{1}{2^{m}}\quad\forall n, m
$$ we obtain
$$
-\frac{1}{2^{n}}+u_{n}\leq u\leq \frac{1}{2^{n}}+u_{n},\quad\forall n.
$$
So, we deduce that for all $n$ and all $z\in \partial \Omega $,
$$
-\frac{1}{2^{n}}+f_{n}(z)\leq
\liminf_{x\in \Omega, x\to z}u(z)\leq \limsup_{x\in \Omega,x\to z
}u(z)\leq \frac{1}{2^{n}}+f_{n}(z) $$ which implies ${\lim_{x\to
z} u(x)}=f(z)$ and completes the proof of Theorem \ref{thDP}.
\hfill$\square$
\begin{remark} \label{remsup} \rm
Using the same techniques as in the proof of Theorem \ref{thDP}
we can show that every increasing and locally bounded
sequence $(u_{n})_{n}$ of supersolutions of \eqref{eq1}
in $\Omega $ is locally bounded in $W^{1,p}(\Omega )$ and that
$u=\lim_{n}u_{n}$ is a supersolution of \eqref{eq1} in
$\Omega $.
\end{remark}
\section{Sheaf property for Superharmonic functions}
\subsection*{The obstacle Problem}
\paragraph{Definition}
Let $f$, $h\in W^{1,p}(\Omega )$ and let
$$
K_{f,h}=\big\{ u\in W^{1,p}(\Omega ):h\leq u\mbox{ a.e. in }
\Omega , u-f\in W_{0}^{1,p}(\Omega )\big\}.
$$
If $f=h$, we denote $K_{f,h}=K_{f}$.
We say that a function $u\in K_{f,h}$ is a solution to the
obstacle problem in $K_{f,h}$ if $$ \int_{\Omega
}\mathcal{A}(.,u,\nabla u).\nabla (v-u)+\int_{\Omega
}\mathcal{B}(.,u,\nabla u)(v-u)\geq 0 $$ whenever $v\in $
$K_{f,h}$.
This function $u$ is called solution of the problem with obstacle $h$
and boundary value $f$.
\begin{remark} \label{rm4.1} \rm
Since $u+\varphi \in K_{f,h}$ for all nonnegative $\varphi \in
W_{0}^{1,p}(\Omega )$, the solution $u$ to the obstacle problem is
always a supersolution of \emph{(\ref{eq1})} in $\Omega $.
Conversely, a supersolution of \emph{(\ref{eq1})} is always a
solution to the obstacle problem in $K_{u}(D)$ for all open
$D\subset \overline{D}\subset \Omega$.
\end{remark}
\begin{theorem} \label{thop}
Let $h$ and $f\ $ be in $W^{1,p}(\Omega )\cap L^{\infty }(\Omega)$.
If $v$ is an upper bounded supersolution of \eqref{sD}
with boundary value $f$ such that $v\geq h$, then there exists
a solution $u$ to the obstacle problem in $K_{f,h}$ with $u\leq v$.
\end{theorem}
\paragraph{Proof.}
As in \cite{Leo}, we introduce the function
$$
g(x,\zeta ,\xi )=\begin{cases}
\widetilde{\mathcal{B}}(x,\zeta ,\xi ) & \mbox{if } \zeta \leq v(x)\\
\widetilde{\mathcal{B}}(x,v,\nabla v) & \mbox{if } \zeta >v(x).
\end{cases}
$$
As in \cite{Hes}, we define the function
$$
\mbox{\bf {a}}(x,\zeta ,\xi )=\begin{cases}
\mathcal{A}(x,\zeta ,\xi ) & \mbox{if } \zeta \leq v(x)\\
\mathcal{A}(x,v,\nabla v) & \mbox{if } \zeta>v(x).
\end{cases}
$$ Note that $\bf {a}$ satisfies the conditions (P1), (P2), and
(P3).
A Lemma in \cite[p.52]{Deu} proves that the map $u\to g(x,u,\nabla
u) $ from $W^{1,p}(\Omega )$ to $L^{p'}(\Omega )$ is bounded and
continuous. Without loss of generality we can assume that $r\geq p-1$.
Let $l=\max \{ q',\frac{p}{p-r}\} -1$, and
define the following penalty term
$$ \gamma (x,s)=[(s-v(x))^{+}]^l \quad \forall x\in \Omega , s\in\mathbb{R}.
$$
Let $M>0$ and consider the map $T:K_{0,h}$ $\to
W^{-1,p'}(\Omega )$ defined by
$$
\langle T(u),w\rangle =\int_{\Omega }\mbox{\bf{a}}(.,u,\nabla
u)\nabla w+\int_{\Omega }g(.,u,\nabla
u)w+M\int_{\Omega }\gamma (.,u)w.
$$
Then for any $u,w\in K_{0,h}$, we have
\begin{gather*}
| \int_{\Omega }g(x,u,\nabla u)w| \leq
c_{1}\| w\| _{l+1}+c_{2}\| \nabla u\|_{p}^{r}\| w\| _{l+1}, \\
| \int_{\Omega}\gamma (x,u)w| \leq c_{3}\| w\|_{l+1}
+c_{4}\| u\| _{l+1}^{l}\| w\| _{l+1},
\end{gather*}
and for each $u\in K_{f,h}-f$ , we have
$$
\int_{\Omega}\gamma (.,u)u\geq c_{5}\| u\| _{l+1}^{l+1}-c_{6}.
$$
An easy computation shows that for $\varepsilon >0$,
\begin{eqnarray*}
(T(u),u) &\geq &(\alpha -c_{2}\varepsilon )\| \nabla u\|
_{p}^p-(c\| u\| _{p}^p+c_{1}\| u\|
_{l+1}^{l+1}+c_{2}c(\varepsilon )\| u\| _{l+1}^{l+1}) \\
&&+Mc_{5}\| u\| _{l+1}^{l+1}-Mc_{6}-c_{1}c_{7}.
\end{eqnarray*}
where $c(\varepsilon )$ is a constant which depends on $\varepsilon $ and
$c>0$. Now, we choose $M$ large to get the operator $T$ coercive.
Since $T$ is bounded , pseudomonotone and continuous, then by a Theorem
in \cite{Ne}, there exists $w\in $ $K_{0,h}$ such that $(T(w),u-w)\geq 0$ for
all $ u\in K_{0,h}$.
Next, we show that $w\leq v$. Since $w-((w-v)\vee 0)\in K_{0,h}$
and since $v$ is a supersolution of (\ref{sD}), it follows that $$
\int_{\{ w>v\} }[\mathcal{A}(.,w,\nabla w)- \mathcal{A}(.,v,\nabla
v)]\nabla (w-v) \leq M\int_{\{ w>v\} }\gamma (.,w)(v-w). $$ Thus
by (P2), $(w-v)^{+}=0$ a.e. in $\Omega $ and hence, $w\leq v$ on
$\Omega $. Finally, if we take $w_{1}=w+f$, we obtain a
supersolution of the obstacle problem $K_{f,h}$. \hfill$\square$
\subsection*{Nonlinear Harmonic Space}
\paragraph{Definition}
Let $V$ be a regular set. For every $f\in C(\partial V)$, we denote by
$H_{V}f$ the solution of the Dirichlet problem with the boundary data $f$.
\begin{proposition}
Let $f$ and $g$ in $C(\partial V)$ be such that $f\leq g$. Then
\begin{description}
\item[i)] $H_{V}f\leq H_{V}g$
\item[ii)] For every $k\geq 0$, we have $H_{V}(k+f)\leq H_{V}(f)+k$
and $H_{V}(f)-k\leq H_{V}(f-k)$.
\end{description}
\end{proposition}
\paragraph{Definition}
Let $U$ be an open set. We denote by $\mathcal{U}(U)$ the set of
all open, regular subsets of $U$ which are relatively compact in
$U$.
We say that a function $u$ is harmonic on $U$, if $u\in C(U)$ and
$u$ is a solution of (\ref{eq1}). We denote by $\mathcal{H}(U)\
$the set of all harmonic functions on $U$. Then,
$$
\mathcal{H}(U)=\big\{ u\in C(U):H_{V}u=u\mbox{ for every }V\in \mathcal{U}
(U)\big\}.
$$
A lower semicontinuous function $u$ is said to be hyperharmonic on
$U$, if
\begin{itemize}
\item $-\infty 0$. Choose a regular open set $V\subset
\overline{V}\subset U$ such that $x\in V$ and $v0$, since $u-\varepsilon \leq
u_{n}\leq u$ $+\varepsilon $ , we get $H_{V}(u)-\varepsilon \leq
u_{n}\leq H_{V}(u)+\varepsilon $ and therefore $H_{V}(u)=u$
\hfill$\square$
\begin{theorem}
Suppose that the conditions in subsection \ref{sDP} are satisfied,
$k_{0}=e=k=0$ and $\alpha \geq p-1$. Then
$(\mathbb{R}^{d},\mathcal{H})$ is a nonlinear Bauer harmonic
space.
\end{theorem}
\paragraph{Proof.}
It is clear that $\mathcal{H}$ is a sheaf of continuous functions
and by Theorem \ref{thDP} there exists a basis of regular sets
stable by intersection. The Bauer convergence property is
fulfilled by Theorem \ref {thcB}. Since $k_{0}=e=k=0$ and $\alpha
\geq p-1$, we have the following form of the Harnack inequality
(e.g. \cite{Ma},\cite{Tr} or \cite{Ser}): For every non empty open
set $U$ in $\mathbb{R}^{d}$, for every constant $M>0$ and every
compact $K$ in $U$, there exists a constant $C=C(K,M)$ such hat
$$
\sup_{K}u\leq C\inf_{K}u
$$
for every $u\in \mathcal{H}^{+}(U)$
with $u\leq M$. It follows that the sheaf $\mathcal{H}$ is non
degenerate. \hfill$\square$
\begin{theorem}
Suppose that the condition of strict monotony holds. Let $u$ $\in $ $%
\mathcal{H}^{\ast }(\Omega )\cap L^{\infty }(\Omega )$. Then $u$ is a
supersolution on $U.$
\end{theorem}
\paragraph{Proof.}
Let $V\subset \overline{V}\subset \Omega $. Let $(\varphi _i)_i$
be an increasing sequence in $C_{c}^{\infty }(\Omega )$ such that
$\ u={\sup_i} \varphi _i$ on $\overline{V}$. Let
$$
K_{\varphi _i}=\big\{ w\in W_{{\rm loc}}^{1,p}(\Omega ):\varphi _i\leq w%
\mbox{, \ }w-\varphi _i\in W_{0}^{1,p}(V)\big\} .
$$
We know by Theorem \ref{thop} that there exists a solution $u_i$ to the
obstacle problem $K_{\varphi _i}$ such that
$\|u_i\|_{\infty }\leq \| \varphi _i\| _{\infty }$. We claim that
$(u_i)_i$ is increasing. In fact $u_i\wedge u_{i+1}\in K_{\varphi _i}$,
then
\begin{eqnarray*}
\int_{\{ u_i>u_{i+1}\}
}(\mathcal{A}(.,u_i,\nabla u_i)-\mathcal{A}(.,u_{i+1},\nabla
u_{i+1}))\nabla (u_{i+1}-u_i) & \\ +\int_{\{
u_i>u_{i+1}\} }(\mathcal{B}(.,u_i,\nabla
u_i)-\mathcal{B}(.,u_{i+1},\nabla u_{i+1}))(u_{i+1}-u_i) &
\geq 0.
\end{eqnarray*}
Hence $\nabla (u_{i+1}-u_i)^{+}=0$ a.e. which yields that
$u_i\leq u_{i+1}$ a.e. in $V$ .
On the other hand, for each $i$ the function $u_i$ is a solution
of (\ref {eq1}) in $D_i:=\{ \varphi _i0$ such that $\varepsilon
\| \psi \|
\leq \inf_{\overline{W}}(u_i-\varphi _i)$. Then, we get $%
u_i+\varepsilon \psi \in K_{\varphi _i}$ and
$$
\int_{W }\mathcal{A}(.,u_i,\nabla u_i).\nabla \psi
+\int_{W }\mathcal{B}(.,u_i,\nabla u_i)\psi =0.
$$
Since
$$
\lim \inf_{x\to y} u(x)\geq u(y)\geq \varphi
_i(y)={\lim_{x\to y} }u_i(x)
$$
for all $y\in \partial D_i$, it yields, by the comparison principle, that
$u\geq u_i$ in $D_i$. Hence $u\geq u_i$ in $D$. Thus
$u={\lim_{i\to \infty } }\varphi _i\leq
\lim_{i\to \infty } u_i\leq u$.
Finally, using Remark \ref{remsup} we complete the proof. \hfill$\square$
\begin{theorem} \label{thm4.45}
Suppose that the condition of strict monotonicity holds. Then
$^{\ast }\mathcal{H}$ is a sheaf.
\end{theorem}
The proof of this theorem is the same as in \cite[Theorem 4.2]{BB}.
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\noindent\textsc{Azeddine Baalal}\\
D\'{e}partement de Math\'{e}matiques et d'Informatique,\\
Facult\'{e} des Sciences A\"{i}n Chock, \\
Km 8 Route El Jadida BP 5366 M\^{a}arif, Casablanca, Maroc.\\
E-mail: baalal@facsc-achok.ac.ma \medskip
\noindent\textsc{Nedra BelHaj Rhouma}\\
Institut Pr\'{e}paratoire aux Etudes d'Ing\'{e}nieurs de Tunis,\\
2 Rue Jawaher Lel Nehru, 1008 Montfleury, Tunis, Tunisie.\\
E-mail: Nedra.BelHajRhouma@ipeit.rnu.tn
\end{document}